A wood-burning stove stands unused in a room where the temperature is `18^(@)C(291K)`. A fire is started inside the stove. Eventually, the temperature of the stove surface reaches a constant `198^(@)C(471K)`, and the room warms to a constant `29^(@)C(302K)`. The stove has an emissivity of `0.900` and a surface area of `3.50m^(2)`. Determine the net radiant power generated by the stove when the stove `(a)` is unheated and has a temperature equal to room temperature and `(b)` has a temperature of `198^(@)C`.

Explanation

Reasoning : The stove emits more radiant power heated than when unheated. In both cases, however, the Stefan-Bolfzmann law can be used to determine the amount of power emitted. Power is energy per unit time or `Q//t`. But in this problem we need to find the net power produced by the stove. The net power is the power the stove emits minus the power the stove absorbs. Then power the stove absorbs comes from the wall, ceiling, and floor of the room, all of which emit radiation.
`(a)` Remembering that temperature must be expressed in kelvins when using the Stefan-Boltzmann law, we find that
Power emitted by unheated `=(Q)/(t)=e sigmaT^(4)A`
`=(0.900)[5.67xx10^(-8)J//(s*m^(2)*K^(4)](291K)^(4)(3.50m^(2))=1280W`
The fact that the unheated stove emits `1280W` of power and yet maintains a constant tempertaures means that the stove also absorbs `1280W` of radiant power from its surroundings. Thus, the net power generated by the unheated stove is zero.
Net power generated by stove at `18^(@)C=underset("Power emitted by stove at" 180^(@)C)(underbrace(1280W))-underset("Power emitted by room at" 180^(@)C"and absorbed by stove")(underbrace(1280W))=0W`
`(b)` The hot stove `(198^(@)C)` or `471K`) emits more radiant power than it absorbs from the cooler room. The radiant power the stove emits is
Power emitted by stove at `198^(@)C=(Q)/(t)=esigmaT^(4)A`
`=(0.900)[5.67xx10^(-8)J//(S*m^(2)*K^(4)](471K)^(4)(3.50m^(2))=8790W`
The radiant power the stove absorbs from the room is identical to the power that the stove would emit at the constant room temperature of `29^(@)C(302K)`. The reasoning here is exactly like that in part `(a)`.
Power emitted by
room at `29^(@)C` and `=(Q)/(t)=esigma^(4)A`
absorbed by stove
`=(0.900)[5.67xx10^(-8J//(s*m^(2)*K^(4))](302K)^(4)(3.50m^(2))=1490W`
The net radiant power the stove produces from the fuel it burn is
Net power
`underset("stove at" 198^(@)C)(generated by)=underset("Power emitted by stove at" 198^(@)C)(underbrace(8790W))-underset("Power emitted by room at" 29^(@)C"and absorbed by stove")(underbrace(1490W))=7300W`